What happens when the first step in a multi-step reaction is not the
rate-limiting step? Consider the reaction between NO and O2 to form NO2,
for example.

2 NO(g) + O2(g) 2 NO2(g)

This reaction involves a two-step mechanism. The first step is a
relatively fast reaction in which a pair of NO molecules combine to form a dimer, N2O2.
The product of this step then undergoes a much slower reaction in which it combines with O2
to form a pair of NO2 molecules.

Step 1:

2 NO N2O2

(fast step)

Step 2:

N2O2 + O2
2 NO2

(slow step)

The net effect of these reactions is the transformation of two NO
molecules and one O2 molecule into a pair of NO2 molecules.

2 NO N2O2

N2O2 + O2 2 NO2

_______________

2 NO + O2
2 NO2

In this reaction, the second step is the rate-limiting step. No matter how
fast the first step takes place, the overall reaction cannot proceed any faster than the
second step in the reaction. As we have seen, the rate of any step in a reaction is
directly proportional to the concentrations of the reactants consumed in that step. The
rate law for the second step in this reaction is therefore proportional to the
concentrations of both N2O2 and O2.

Step 2:

Rate2nd = k(N2O2)(O2)

Because the first step in the reaction is much faster, the overall rate of
reaction is more or less equal to the rate of this rate-limiting step.

Rate k(N2O2)(O2)

This rate law is not very useful because it is difficult to measure the
concentrations of intermediates, such as N2O2, that are
simultaneously formed and consumed in the reaction. It would be better to have an equation
that related the overall rate of reaction to the concentrations of the original reactants.

Let's take advantage of the fact that the first step in this reaction is
reversible.

Step 1:

2 NO N2O2

The rate of the forward reaction in this step depends on the concentration
of NO raised to the second power.

Step 1:

Rateforward = kf(NO)2

The rate of the reverse reaction depends only on the concentrations of N2O2.

Step 1:

Ratereverse = kr(N2O2)

Because the first step in this reaction is very much faster than the
second, the first step should come to equilibrium. When that happens, the rate of the
forward and reverse reactions for the first step are the same.

kf(NO)2 = kr(N2O2)

Let's rearrange this equation to solve for one of the terms that appears
in the rate law for the second step in the reaction.

(N2O2) = (kf /kr)
(NO)2

Substituting this equation into the rate law for the second step gives the
following result.

Rate2nd = k (kf /kr) (NO)2(O2)

Since k, kf, and kr are all
constants, they can be replaced by a single constant, k', to give the experimental
rate law for this reaction described in Exercise 22.6.

There is a simple relationship between the equilibrium constant for a
reversible reaction and the rate constants for the forward and reverse reactions if the
mechanism for the reaction involves only a single step. To understand this
relationship, let's turn once more to a reversible reaction that we know occurs by a
one-step mechanism.

ClNO2(g) + NO(g) NO2(g) + ClNO(g)

The rate of the forward reaction is equal to a rate constant for this
reaction, kf, times the concentrations of the reactants, ClNO2
and NO.

Rateforward = kf(ClNO2)(NO)

The rate of the reverse reaction is equal to a second rate constant, kr,
times the concentrations of the products, NO2 and ClNO.

Ratereverse = kr(NO2)(ClNO)

This system will reach equilibrium when the rate of the forward reaction
is equal to the rate of the reverse reaction.

Rateforward = Ratereverse

Substituting the rate laws for the forward and reverse reactions when the
system is at equilibrium into this equation gives the following result.

kf[NO][ClNO2] = kr[ClNO][NO2]

This equation can be rearranged to give the equilibrium constant
expression for the reaction.

Thus, the equilibrium constant for a one-step reaction
is equal to the forward rate constant divided by the reverse rate constant.

Practice Problem 7:

The rate constants for
the forward and reverse reactions in the following equilibrium have been measured.
At 25°C, kf is 7.3 x 103 liters per mole-second and kris 0.55 liters per mole-second. Calculate the equilibrium constant for this
reaction:

The rate law for a reaction can be determined by studying what happens to
the initial instantaneous rate of reaction when we start with different initial
concentrations of the reactants. To show how this is done, let's determine the rate law
for the decomposition of hydrogen peroxide in the presence of the iodide ion.

2 H2O2(aq)

2 H2O(aq) + O2(g)

Data on initial instantaneous rates of reaction for five experiments run
at different initial concentrations of H2O2 and the I-
ion are given in the table below.

Rate of Reaction Data for the Decomposition of H2O2
in the Presence of the I- Ion

Initial (H2O2) (M)

Initial (I-) (M)

Initial Instantaneous
Rate of Reaction (M/s)

Trial 1:

1.0 x 10-2

2.0 x 10-3

2.3 x 107

Trial 2:

2.0 x 10-2

2.0 x 10-3

4.6 x 107

Trial 3:

3.0 x 10-2

2.0 x 10-3

6.9 x 107

Trial 4:

1.0 x 10-2

4.0 x 10-3

4.6 x 107

Trial 5:

1.0 x 10-2

6.0 x 10-3

6.9 x 107

The only difference between the first three trials is the initial
concentration of H2O2. The difference between Trial 1 and Trial 2 is
a two-fold increase in the initial H2O2 concentration, which leads
to a two-fold increase in the initial rate of reaction.

The difference between Trial 1 and Trial 3 is a three-fold increase in the
initial H2O2 concentration, which produces a three-fold increase in
the initial rate of reaction.

The only possible conclusion is that the rate of reaction is directly
proportional to the H2O2 concentration.

Experiments 1, 4, and 5 were run at the same initial concentration of H2O2
but different initial concentrations of the I- ion. When we compare Trials 1
and 4 we see that doubling the initial I- concentration leads to a twofold
increase in the rate of reaction.

Trials 1 and 5 show that tripling the initial I- concentration
leads to a three-fold increase in the initial rate of reaction. We therefore conclude that
the rate of the reaction is also directly proportional to the concentration of the I-
ion.

The results of these experiments are consistent with a rate law for this
reaction that is first-order in both H2O2 and I-.

The rate law for a reaction is a useful way of probing the mechanism of a
chemical reaction but it isn't very useful for predicting how much reactant remains in
solution or how much product has been formed in a given amount of time. For these
calculations, we use the integrated form of the rate law.

Let's start with the rate law for a reaction that is first-order in the
disappearance of a single reactant, X.

When this equation is rearranged and both sides are integrated we get the
following result.

Integrated form of the first-order rate law:

In this equation, (X) is the concentration of X at any
moment in time, (X)0 is the initial concentration of this reagent, k
is the rate constant for the reaction, and t is the time since the reaction
started.

To illustrate the power of the integrated form of the rate law for a
reaction, let's use this equation to calculate how long it would take for the 14C
in a piece of charcoal to decay to half of its original concentration. We will start by
noting that 14C decays by first-order kinetics with a rate constant of 1.21 x
10-4 yr-1.

The integrated form of this rate law would be written as follows.

We are interested in the moment when the concentration of 14C
in the charcoal is half of its initial value.

Substituting this relationship into the integrated form of the rate law
gives the following equation.

We now simplify this equation

and then solve for t.

It therefore takes 5730 years for half of the 14C in the sample
to decay. This is called the half-life of 14C. In general, the half-life
for a first-order kinetic process can be calculated from the rate constant as follows.

Let's now turn to the rate law for a reaction that is second-order in a
single reactant, X.

The integrated form of the rate law for this reaction is written as
follows.

Integrated form of the second-order rate law:

Once again, (X) is the concentration of X at any moment in
time, (X)0 is the initial concentration of X, k is the
rate constant for the reaction, and t is the time since the
reaction started.

Practice Problem 9:

Acetaldehyde, CH3CHO,
decomposes by second-order kinetics with a rate constant of 0.334 M-1s-1
at 500°C. Calculate the amount of time it would take for 80% of the acetaldehyde to
decompose in a sample that has an initial concentration of 0.00750 M.

The half-life of a second-order reaction can be calculated from the
integrated form of the second-order rate law.

We start by asking: "How long it would take for the concentration of X
to decay from its initial value, (X)0, to a value half as large?"

The first step in simplifying this equation involves multiplying the top
and bottom halves of the first term by 2.

Subtracting one term on the left side of this equation from the other
gives the following result.

We can now solve this equation for the half-life of the reaction.

There is an important difference between the equations for calculating the
half-life of first order and second-order reactions. The half-life of a first-order
reaction is a constant, which is proportional to the rate constant for the reaction.

first-order reaction:

The half-life for a second-order reaction is inversely proportional to
both the rate constant for the reaction and the initial concentration of the reactant that
is consumed in the reaction.

second-order reaction:

Discussions of the half-life of a reaction are therefore usually confined
to first-order processes.

The integrated form of the rate laws for first- and second-order reactions
provides another way of determining the order of a reaction. We can start by assuming, for
the sake of argument, that the reaction is first-order in reactant X.

Rate = k(X)

We then test this assumption by checking concentration versus time data
for the reaction to see whether they fit the first-order rate law.

To see how this is done, let's start by rearranging the integrated form of
the first-order rate law as follows.

ln (X) - ln (X)0 = - kt

We then solve this equation for the natural logarithm of the concentration
of X at any moment in time.

ln (X) = ln (X)0 - kt

This equation contains two variablesln (X) and tand two constantsln (X)0 and k. It can therefore be set up in terms of
the equation for a straight line.

y = mx + b

ln (X) = -kt + ln (X)0

If the reaction is first-order in X, a plot of the natural
logarithm of the concentration of X versus time will be a straight line with a
slope equal to -k, as shown in the figure below.

If the plot of ln (X) versus time is not a straight line, the
reaction can't be first-order in X. We therefore assume, for the sake of argument,
that it is second-order in X.

Rate = k(X)2

We then test this assumption by checking whether the experimental data fit
the integrated form of the second-order rate law.

This equation contains two variables(X) and tand two constants(X)0 and k. Thus, it also can be set up in terms of
the equation for a straight line.

y = mx + b

If the reaction is second-order in X, a plot of the reciprocal of
the concentration of X versus time will be a straight line with a slope equal to k,
as shown in the figure below.

Use
the experimental data found in the table at the beginning of this lesson to determine
whether the reaction between phenolphthalein (PHTH) and the OH- ion is a
first-order or a second-order reaction.

What about reactions that are first-order in two reactants, X and Y, and
therefore second-order overall?

Rate = k(X)(Y)

A plot of 1/(X) versus time won't give a straight line because the
reaction is not second-order in X. Unfortunately, neither will a plot of ln (X)
versus time, because the reaction is not strictly first-order in X. It is
first-order in both X and Y.

One way around this problem is to turn the reaction into one that is pseudofirst-order
by making the concentration of one of the reactants so large that it is effectively
constant. The rate law for the reaction is still first-order in both reactants. But the
initial concentration of one reactant is so much larger than the other that the rate of
reaction seems to be sensitive only to changes in the concentration of the reagent present
in limited quantities.

Assume, for the moment, that the reaction is studied under conditions for
which there is a large excess of Y. If this is true, the concentration of Y
will remain essentially constant during the reaction. As a result, the rate of the
reaction will not depend on the concentration of the excess reagent. Instead, it will
appear to be first order in the other reactant, X. A plot of ln (X) versus
time will therefore give a straight line.

Rate = k'(X)

If there is a large excess of X, the reaction will appear to be
first-order in Y. Under these conditions, a plot of log (Y) versus time will
be linear.

Rate = k'(Y)

The value of the rate constant obtained from either of these equationsk'won't be the actual rate constant for the reaction. It
will be the product of the rate constant for the reaction times the concentration of the
reagent that is present in excess.

In our discussion of acid-base equilibria, we argued that the
concentration of water is so much larger than any other component of these solutions that
we can build it into the equilibrium constant expression for the reaction.

We now understand why this is done. Because the concentration of water is
so large, the reaction between an acid or a base and water is a pseudo-first-order
reaction that only depends on the concentration of the acid or base.